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Physics 486 Discussion Solution Week 1
1: We won’t discuss Bohr’s model of the atom in lecture, but it is historically interesting. It was the first phenomenologically correct model of the hydrogen atom, and led to de Broglie’s wave hypothesis. The idea is very simple (many of the best ones are ...): Consider an electron in a circular orbit around a proton. (Elliptical orbits work also, but the math is messier.) The “action” (the integral, !
integral multiple of some constant (call it K ).^ #^ r^ p^ "^ d r^ l) is required to be an a: Remember that the Coulomb force is, !
electron when it is in a circular orbit of radius,^^ F^ =^4 "#^10 r^ r. Assume that proton is stationary.^ e^22. Calculate the momentum of the Use F = mv^2 / r ⇒ !
p^2 = mrF = (^4) " m # 0^ e r^2 b: Calculate the action (usually denoted by the symbol, S ). S = 2π rp = !
" me # 02 r c: Requiring that S = nK , calculate the radius of the orbit. r = !
n #^2 meK^22 " 0 d: We now know the potential energy, !
using the virial theorem: = - /2. (Do you know the virial theorem? It’s^^ V^ =^ "^4 #$^10^ e^ r^2. We can calculate the total energy sometimes very useful.) E = !
" (^8) #$^10^ e r^2 = (^8) $ 02 men (^24) K 2 e: You should have obtained that E = - 13. 6 eV, to obtain a numerical value for r ∝ n^2 , and E ∝ K 1/. You should obtain an amazing result. n^2. Now use the fact that for n = 1, Using SI units: K = 6.626× 10 - (^34) J.s ( ε m (^0) i.e. (^) e = 8.834 = 9.109, Planck’s constant).×× 1010 -^12 - 31 F/m (^) kg e 1 eV = 1.602 == 1.602× (^10) × 10 -^19 - C (^19) J: When Bohr obtained this result in 1913, the coincidence was a big puzzle to be solved.
2: If we are to take seriously the concept of photons, we should see the particle nature when we scatter light from electrons. This is called Compton scattering. Let’s compare the scattering of waves and particles from (initially stationary) electrons:
a: Suppose the incoming wave has wavelength, outgoing wave? Use classical E&M ( i.e. , P436) here. λ 0. What is the wavelength, λ 1 , of the In classical E&M, the electron oscillates with the freq radiates an outgoing wave with that frequency, so λ 1 =uency of the incoming wave. It λ 0. b: Suppose the incoming photon has momentum, outgoing photon? You’ll need to use special relativity, because photons are massl p 0. What is the momentum, p 1 , of theess. This is a relativistic billiard ball problem. Use conservation of energy and momentum. Momentum: !
p 00 == pp 11 cossin " " %+ pp sin cos # # $$ p ( 1 p sin 0 % " p = 1 cos p sin ") # = p cos # Square and add ( φ drops out): !
p 02 + p 12 " 2 p 0 p 1 cos # = p^2 (1) Energy: !
p 0 c + mc^2 = p 1 c + ( mc^2 )^2 + ( pc )^2 ⇒ ! (1) and (2) both =^ (^ p^0 "^ p^1 ) p^22 , so set them equal (+^2 mc (^ p^0 "^ p^1 )^ =^ pp^2 no longer appears) and solve: (2)
!
p^11 "^ p^10 =^1 "^ mc cos^ # c: Use the QM relation, Compare this answer with that to part a. For what range of wavelengths d p = h /λ, to relate the incoming and outgoing wavelengths in part b.o they agree? Using 1/p = λ/h, we have: !
" 1 # " 0 = (^) mch ( 1 # cos $ ) h photon’s wavelength, /mc ≡ λC = 3.86× 10 -^13 λ^ m is called the Compton wavelength of the electron. When the= 2π λC, its energy = mc (^2). As long as λ 0 >> λC, we have λ 1 ≈ λ 0.
λ 0 λ^1 p 0^ p^1 θ φ p
